Given an array arr[] of N integers, the task is to print all the element of Longest Dividing Subsequence formed by the given array. If we have more than one sequence with maximum length then print all of them.
Examples:
Input: arr[] = { 2, 11, 16, 12, 36, 60, 71 }
Output:
2 12 36
2 12 60
Explanation:
There are two subsequence with maximum length 3.
1. 2, 12, 36
2. 2, 12, 60Input: arr[] = { 2, 4, 16, 32, 64, 60, 12 };
Output:
2 4 16 32 64
Explanation:
There is only one subsequence with maximum length 5.
1. 2 4 16 32 64
Approach:
- Declare a 2D array LDS[][] which stores the Longest Dividing Subsequence.
- Traverse the given array arr[] and find the longest dividing subsequence till current element by using the approach discussed in this article.
- Traverse the given LDS[][] array, and print all the elements of the sequence with maximum length.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include "bits/stdc++.h" using namespace std; // Function to print LDS[i] element void printLDS(vector<int>& Max) { // Traverse the Max[] for (auto& it : Max) { cout << it << ' '; } } // Function to construct and print Longest // Dividing Subsequence void LongestDividingSeq(int arr[], int N) { // 2D vector for storing sequences vector<vector<int> > LDS(N); // Push the first element to LDS[][] LDS[0].push_back(arr[0]); // Interate over all element for (int i = 1; i < N; i++) { // Loop for every index till i for (int j = 0; j < i; j++) { // if current elements divides // arr[i] and length is greater // than the previous index, then // insert the current element // to the sequences LDS[i] if ((arr[i] % arr[j] == 0) && (LDS[i].size() < LDS[j].size() + 1)) LDS[i] = LDS[j]; } // L[i] ends with arr[i] LDS[i].push_back(arr[i]); } int maxLength = 0; // LDS stores the sequences till // each element of arr[] // Traverse the LDS[][] to find the // the maximum length for (int i = 0; i < N; i++) { int x = LDS[i].size(); maxLength = max(maxLength, x); } // Print all LDS with maximum length for (int i = 0; i < N; i++) { // Find size int size = LDS[i].size(); // If size = maxLength if (size == maxLength) { // Print LDS printLDS(LDS[i]); cout << '\n'; } } } // Driver Code int main() { int arr[] = { 2, 11, 16, 12, 36, 60, 71 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call LongestDividingSeq(arr, N); return 0; } |
Python3
# Python3 program for the above approach # Function to print LDS[i] element def printLDS(Max): # Traverse the Max[] for it in Max: print(it, end = " ") # Function to construct and print # Longest Dividing Subsequence def LongestDividingSeq(arr, N): # 2D vector for storing sequences LDS = [[] for i in range(N)] # Push the first element to LDS[][] LDS[0].append(arr[0]) # Interate over all element for i in range(1, N): # Loop for every index till i for j in range(i): # If current elements divides # arr[i] and length is greater # than the previous index, then # insert the current element # to the sequences LDS[i] if ((arr[i] % arr[j] == 0) and (len(LDS[i]) < len(LDS[j]) + 1)): LDS[i] = LDS[j].copy() # L[i] ends with arr[i] LDS[i].append(arr[i]) maxLength = 0 # LDS stores the sequences till # each element of arr[] # Traverse the LDS[][] to find # the maximum length for i in range(N): x = len(LDS[i]) maxLength = max(maxLength, x) # Print all LDS with maximum length for i in range(N): # Find size size = len(LDS[i]) # If size = maxLength if (size == maxLength): # Print LDS printLDS(LDS[i]) print() # Driver Code arr = [ 2, 11, 16, 12, 36, 60, 71 ] N = len(arr) # Function call LongestDividingSeq(arr, N); # This code is contributed by ANKITKUMAR34 |
2 12 36 2 12 60
Time Complexity: O(N2)
Auxiliary Space: O(N2)
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